Timeline for When are these base spaces isomorphic?
Current License: CC BY-SA 4.0
10 events
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| Aug 10 at 10:46 | comment | added | Robert Bryant | @NickL: Yes, that's true. The example I gave is known as the "Klein correspondence", but the compact version replaces $\mathrm{Sp}(4,\mathbb{R})$ with its maximal compact $M=\mathrm{U}(2)$ and then divides by two different circle subgroups of the diagonal matrices. In fact, there are countably many distinct such circle subgroups of $\mathrm{U}(2)$ that give non-homeomorphic quotients. The examples I gave are just two of them. Dividing $\mathrm{SU}(3)$ by its different circle subgroups gives the $7$-dimensional Aloff-Wallach examples, some of which are homemorphic but not diffeomorphic. | |
| Aug 9 at 23:05 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Aug 9 at 19:16 | comment | added | Kevin Casto | A bit more generally yet, given two fiber bundles $f_1: E_1 \to B$ and $f_2: E_2 \to B$, the total spaces of the pullback bundles $f_1^*(E_2)$ and $f_2^*(E_1)$ are diffeomorphic, just by swapping the two two factors in the definition of the pullback. So if the fibers of the two bundles were also diffeomorphic (in other words, two bundles with the same fiber and base), then the two pullbacks give you two bundles with diffeomorphic total spaces and fibers but different bases ($E_1$ and $E_2$ respectively). | |
| Aug 9 at 16:57 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Aug 9 at 10:42 | comment | added | Nick L | I realised my answer was essentially a version of yours hence I delete and leave as a comment here. More generally: Let $X$ be a fibre bundle over a base $B$ with fiber $F$. Then $X \times F$ is a fiber bundle over $X$ and $B \times F$ with fiber $F$. So for example $\mathbb{CP}^3 \times S^2$ is a fibre bundle over $S^4 \times S^2$ and $\mathbb{CP}^3$ with fibre $S^2$. Also $S^3 \times S^1$ is a fiber bundle over $S^2 \times S^1 $ and $S^3$, I think Robert Bryant's answer should be obtained by quotienting this example by an involution. | |
| Aug 9 at 10:24 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Aug 9 at 9:03 | comment | added | HenrikRüping | If you want to reduce the dimensions further, you could just take $M=\mathbb{R}^4_{ex}\times \mathbb{R}\cong \mathbb{R}^5$ and the two projections to $\mathbb{R}^4_{ex}$ and $\mathbb{R}^4$ with fibers $\mathbb{R}$. | |
| Aug 9 at 8:57 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Aug 9 at 8:33 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Aug 9 at 8:26 | history | answered | Francesco Polizzi | CC BY-SA 4.0 |